Question

$\large\huge\color{green} \sf{\dfrac{d}{dx} \left(\dfrac{((sinx {)}^{sinx {} } ) ^{ ^{sinx} } }{ ln(x) } \right) }$
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Answer 1

Step-by-step explanation:

→Use Quotient Rule to find the derivative of $\sf\frac{\sin^{\sin^{\sin{x}}x}x}{\ln{x}}$. The quotient rule states that $\sf(\frac{f}{g})'=\frac{f'g-fg'}{{g}^{2}}$ .

$\tt\red{\implies\frac{\ln{x}(\frac{d}{dx} \sin^{\sin^{\sin{x}}x}x)-\sin^{\sin^{\sin{x}}x}x(\frac{d}{dx} \ln{x})}{{\ln{x}}^{2}}}$

→Use Chain Rule on $\frac{d}{dx} \sin^{\sin^{\sin{x}}.$ Let $\sf u=\sin^{\sin{x}}x\ln{(\sin{x})}.$ The derivative of $\sf{u}^{u}$ is u lnu.

$\tt\pink{\frac{\implies\ln{x}\sin^{\sin^{\sin{x}}x}x(\frac{d}{dx} \sin^{\sin{x}}x\ln{(\sin{x})})-\sin^{\sin^{\sin{x}}x}x(\frac{d}{dx} \ln{x})}{{\ln{x}}^{2}}}$

$\tt\orange{\implies\frac{\ln{x}\sin^{\sin^{\sin{x}}x}x(\sin^{\sin{x}}x(\cos{x}\ln{(\sin{x})}+\cos{x})\ln{(\sin{x})}+\sin^{\sin{x}-1}x\cos{x})-\frac{\sin^{\sin^{\sin{x}}x}x}{x}}{{\ln{x}}^{2}}}$

Answer 2

Question:-

$\large\huge\color{blue} \tt{\dfrac{d}{dx} \left(\dfrac{((sinx {)}^{sinx {} } ) ^{ ^{sinx} } }{ ln(x) } \right) }$

$\huge{\large{\underline{\large \sf{Solution}}}}$

Apply the quotient rule:-

$\rm :\implies{(\dfrac{d}{dx} \: f(x)) g(x) - f(x) \: (\dfrac{d}{dx} g(xI))}$

$\rm :\implies{f(x) - sin(x)}$

$\rm :\implies{g(x) - In(x)}$

$\quad\sf :\longmapsto{\dfrac{(\dfrac{d}{dx} sin(x)) In (x) - sin(x) (\dfrac{d}{dx} in (x))}{In(x)^{2}}}$

$\qquad\sf :\longmapsto{(cos(x) In (x) - \dfrac{sin(x)}{x}).\dfrac{I}{In(x)^{2}}}$

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HOPES IT HELPS!:)